1

© 2007 The McGraw-Hill Companies, Inc. All rights reserved.

V

ector Mechanics for Engineers: Statics

Eighth

Edition

3 - 1

How to prepare for the final

• The final will be based on Chapters 6, 7, 8, and sections 10.1-10.5. It will be

three-hour, take-home, open-textbook and open-notes exam.

• Read “Review and Summary” after each Chapter. Brush up on topics that are

not familiar.

• Make sure you know how to solve HW problems and sample problems.

Additional review problems for the final will be posted on the web.

• Review important tables/formulae from the book (such as supports and their

reactions) so that you can use them easily.

• Remember, the correct reasoning and an error in computation will get you most

of the points. However, the right answer with no explanation will get you no

points, unless the problem specifically asks for an answer only.

• Do not forget about the honor code. Carefully read the instructions on the front

page of the final. You cannot discuss anything about the final until after the due

date.

• The rest of this document is a brief summary of important topics we have learned

in the second half of the term.

© 2007 The McGraw-Hill Companies, Inc. All rights reserved.

V

ector Mechanics for Engineers: Statics

Eighth

Edition

3 - 2

Analysis of Trusses by the Method of Joints

• Dismember the truss and create a freebody

diagram for each member and pin.

• The two forces exerted on each member are

equal, have the same line of action, and

opposite sense.

• Forces exerted by a member on the pins or

joints at its ends are directed along the member

and equal and opposite.

• Conditions of equilibrium on the pins provide

2n equations for 2n unknowns. For a simple

truss, 2n = m + 3. May solve for m member

forces and 3 reaction forces at the supports.

• Conditions for equilibrium for the entire truss

provide 3 additional equations which are not

independent of the pin equations.

2

© 2007 The McGraw-Hill Companies, Inc. All rights reserved.

V

ector Mechanics for Engineers: Statics

Eighth

Edition

3 - 3

Analysis of Trusses by the Method of Sections

• When the force in only one member or the

forces in a very few members are desired, the

method of sections works well.

• To determine the force in member BD, pass a

section through the truss as shown and create

a free body diagram for the left side.

• With only three members cut by the section,

the equations for static equilibrium may be

applied to determine the unknown member

forces, including F

BD

.

© 2007 The McGraw-Hill Companies, Inc. All rights reserved.

V

ector Mechanics for Engineers: Statics

Eighth

Edition

3 - 4

Machines

• Machines are structures designed to transmit

and modify forces. Their main purpose is to

transform input forces into output forces.

• Given the magnitude of P, determine the

magnitude of Q.

• Create a free-body diagram of the complete

machine, including the reaction that the wire

exerts.

• The machine is a nonrigid structure. Use

one of the components as a free-body.

• Taking moments about A,

P

b

a

QbQaPM

A

=−==

∑

0

3

© 2007 The McGraw-Hill Companies, Inc. All rights reserved.

V

ector Mechanics for Engineers: Statics

Eighth

Edition

3 - 5

Shear and Bending Moment Diagrams

• Variation of shear and bending

moment along beam may be

plotted.

• Determine reactions at

supports.

• Cut beam at C and consider

member AC,

22 PxMP

V

+

=

+

=

• Cut beam at E and consider

member EB,

( )

22 xLPMP

V

−

+

=

−

=

• For a beam subjected to

concentrated loads

, shear is

constant between loading points

and moment varies linearly.

© 2007 The McGraw-Hill Companies, Inc. All rights reserved.

V

ector Mechanics for Engineers: Statics

Eighth

Edition

3 - 6

Relations Among Load, Shear, and Bending Moment

• Relations between load and shear:

(

)

w

x

V

dx

dV

x

wVVV

x

−=

Δ

Δ

=

=

Δ

−

Δ

+

−

→

Δ

0

lim

0

( )

curve loadunder area−=−=−

∫

D

C

x

x

CD

dxwVV

• Relations between shear and bending moment:

( )

( )

VxwV

x

M

dx

dM

x

xwxVMMM

x

x

=Δ−=

Δ

Δ

=

=

Δ

Δ+Δ−−Δ+

→

Δ

→

Δ

2

1

00

limlim

0

2

( )

curveshear under area==−

∫

D

C

x

x

CD

dxVMM

4

© 2007 The McGraw-Hill Companies, Inc. All rights reserved.

V

ector Mechanics for Engineers: Statics

Eighth

Edition

3 - 7

Relations Among Load, Shear, and Bending Moment

• Reactions at supports,

2

wL

RR

BA

==

• Shear curve,

⎟

⎠

⎞

⎜

⎝

⎛

−=−=−=

−=−=−

∫

x

L

wwx

wL

wxVV

wxdxwVV

A

x

A

22

0

• Moment curve,

( )

⎟

⎠

⎞

⎜

⎝

⎛

===

−=

⎟

⎠

⎞

⎜

⎝

⎛

−=

=−

∫

∫

0at

8

22

2

max

2

0

0

V

dx

dM

M

wL

M

xxL

w

dxx

L

wM

VdxMM

x

x

A

© 2007 The McGraw-Hill Companies, Inc. All rights reserved.

V

ector Mechanics for Engineers: Statics

Eighth

Edition

3 - 8

Sample Problem 7.4

Draw the shear and bending-

moment diagrams for the beam

and loading shown.

SOLUTION:

• Taking entire beam as a free-body, determine

reactions at supports.

• With uniform loading between D and E, the

shear variation is linear.

• Between concentrated load application

points, and shear is

constant.

0

=

−

=

wdxdV

• Between concentrated load application

points, The change

in moment between load application points is

equal to area under shear curve between

points.

.constant

=

=

VdxdM

• With a linear shear variation between D

and E, the bending moment diagram is a

parabola.

5

© 2007 The McGraw-Hill Companies, Inc. All rights reserved.

V

ector Mechanics for Engineers: Statics

Eighth

Edition

3 - 9

Sample Problem 7.4

• Between concentrated load application

points, The change

in moment between load application points is

equal to area under the shear curve between

points.

.constant

=

=

VdxdM

• With a linear shear variation between D

and E, the bending moment diagram is a

parabola.

048

ftkip 48140

ftkip 9216

ftkip 108108

=+=−

⋅−=−=−

⋅+=−=−

⋅

+

=

+

=

−

E

D

E

DCD

CBC

BAB

MMM

MMM

MMM

MMM

© 2007 The McGraw-Hill Companies, Inc. All rights reserved.

V

ector Mechanics for Engineers: Statics

Eighth

Edition

3 - 10

Cables With Concentrated Loads

• Consider entire cable as free-body. Slopes of

cable at A and B are not known - two reaction

components required at each support.

• Four unknowns are involved and three

equations of equilibrium are not sufficient to

determine the reactions.

• For other points on cable,

2

yields0

2

yM

C

=

∑

yxyx

θθFF, yield 0,0

=

=

∑

∑

捯湳瑡湴捯c

=

=

=

x

x

ATT

θ

• Additional equation is obtained by

considering equilibrium of portion of cable

AD and assuming that coordinates of point D

on the cable are known. The additional

equation is

.0

∑

=

D

M

6

© 2007 The McGraw-Hill Companies, Inc. All rights reserved.

V

ector Mechanics for Engineers: Statics

Eighth

Edition

3 - 11

Cables With Distributed Loads

• For cable carrying a distributed load:

a) cable hangs in shape of a curve

b) internal force is a tension force directed along

tangent to curve.

• Consider free-body for portion of cable extending

from lowest point C to given point D. Forces are

horizontal force T

0

at C and tangential force T at D.

• From force triangle:

0

22

0

0

tan

sincos

T

W

WTT

WTTT

=+=

==

θ

θθ

• Horizontal component of T is uniform over cable.

• Vertical component of T is equal to magnitude of W

measured from lowest point.

• Tension is minimum at lowest point and maximum

at A and B.

© 2007 The McGraw-Hill Companies, Inc. All rights reserved.

V

ector Mechanics for Engineers: Statics

Eighth

Edition

3 - 12

Parabolic Cable

• Consider a cable supporting a uniform, horizontally

distributed load, e.g., support cables for a

suspension bridge.

• With loading on cable from lowest point C to a

point D given by internal tension force

magnitude and direction are

,wxW

=

0

222

0

tan

T

wx

xwTT =+= θ

• Summing moments about D,

0

2

:0

0

=−=

∑

yT

x

wxM

D

0

2

2T

wx

y =

or

The cable forms a parabolic curve.

7

© 2007 The McGraw-Hill Companies, Inc. All rights reserved.

V

ector Mechanics for Engineers: Statics

Eighth

Edition

3 - 13

Catenary

• Consider a cable uniformly loaded along the cable

itself, e.g., cables hanging under their own weight.

• With loading on the cable from lowest point C to a

point D given by the internal tension force

magnitude is

w

T

cscwswTT

0

22222

0

=+=+=

,

wsW

=

• To relate horizontal distance x to cable length s,

c

x

cs

c

s

c

csq

ds

x

csq

ds

T

T

dsdx

s

sinhandsinh

coscos

1

0

22

22

0

==

+

=

+

===

−

∫

θθ

© 2007 The McGraw-Hill Companies, Inc. All rights reserved.

V

ector Mechanics for Engineers: Statics

Eighth

Edition

3 - 14

Catenary

• To relate x and y cable coordinates,

c

x

cy

c

c

x

cdx

c

x

cy

dx

c

x

dx

c

s

dx

T

W

dxdy

x

cosh

coshsinh

sinhtan

0

0

=

−==−

====

∫

θ

which is the equation of a catenary.

8

© 2007 The McGraw-Hill Companies, Inc. All rights reserved.

V

ector Mechanics for Engineers: Statics

Eighth

Edition

3 - 15

The Laws of Dry Friction. Coefficients of Friction

• Four situations can occur when a rigid body is in contact with

a horizontal surface:

• No friction,

(P

x

= 0)

• No motion,

(P

x

< F

m

)

• Motion impending,

(P

x

= F

m

)

• Motion,

(P

x

> F

m

)

© 2007 The McGraw-Hill Companies, Inc. All rights reserved.

V

ector Mechanics for Engineers: Statics

Eighth

Edition

3 - 16

Wedges

• Wedges - simple

machines used to raise

heavy loads.

• Friction prevents wedge

from sliding out.

• Want to find minimum

force P to raise block.

• Block as free-body

0

:0

0

:0

21

21

=+−−

=

=+−

=

∑

∑

NNW

F

NN

F

s

y

s

x

μ

μ

or

0

21

=++ WRR

( )

( )

06sin6cos

:0

0

6sin6cos

:0

32

32

=°−°+−

=

=+

°−°−−

=

∑

∑

s

y

ss

x

NN

F

P

NN

F

μ

μμ

• Wedge as free-body

or

0

32

=+− RRP

9

© 2007 The McGraw-Hill Companies, Inc. All rights reserved.

V

ector Mechanics for Engineers: Statics

Eighth

Edition

3 - 17

Square-Threaded Screws

• Square-threaded screws frequently used in jacks, presses, etc.

Analysis similar to block on inclined plane. Recall friction

force does not depend on area of contact.

• Thread of base has been “unwrapped” and shown as straight

line. Slope is 2

π

r horizontally and lead L vertically.

• Moment of force Q is equal to moment of force P.

rPaQ =

• Impending motion

upwards. Solve for

Q

.

• Self-locking, solve

for

Q

to lower load.

,

θ

φ

>

s

• Non-locking, solve

for Q to hold load.

,

θ

φ

>

s

© 2007 The McGraw-Hill Companies, Inc. All rights reserved.

V

ector Mechanics for Engineers: Statics

Eighth

Edition

3 - 18

Journal Bearings. Axle Friction

• Angle between R and

normal to bearing

surface is the angle of

kinetic friction ϕ

k

.

k

k

Rr

R

r

M

μ

φ

≈

= sin

• May treat bearing

reaction as force-

couple system.

• For graphical solution,

R must be tangent to

circle of friction.

k

kf

r

rr

μ

φ

≈

=

sin

10

© 2007 The McGraw-Hill Companies, Inc. All rights reserved.

V

ector Mechanics for Engineers: Statics

Eighth

Edition

3 - 19

Belt Friction

• Relate T

1

and T

2

when belt is about to slide to right.

• Draw free-body diagram for element of belt

( )

0

2

cos

2

cos:0 =Δ−

Δ

−

Δ

Δ+=

∑

NTTTF

sx

μ

θ

θ

( )

0

2

sin

2

sin:0 =

Δ

−

Δ

Δ+−Δ=

∑

θ

θ

TTTNF

y

• Combine to eliminate ΔN, divide through by Δθ,

(

)

2

2sin

22

cos

θ

θ

μ

θ

θ

Δ

Δ

⎟

⎠

⎞

⎜

⎝

⎛

Δ

+−

Δ

Δ

Δ

T

T

T

s

• In the limit as Δθgoes to zero,

0=− T

d

dT

s

μ

θ

• Separate variables and integrate from

β

θ

θ

== to0

βμ

βμ

s

e

T

T

T

T

s

==

1

2

1

2

orln

© 2007 The McGraw-Hill Companies, Inc. All rights reserved.

V

ector Mechanics for Engineers: Statics

Eighth

Edition

3 - 20

Principle of Virtual Work

• Imagine a small virtual displacement of a particle which

is acted upon by several forces.

• The corresponding virtual work,

(

)

r

R

rFFFrFrFrFU

δ

δδδδδ

⋅

=

⋅++=⋅+⋅+⋅=

321321

Principle of Virtual Work:

• A particle is in equilibrium if and only if the total virtual

work of forces acting on the particle is zero for any virtual

displacement.

• A rigid body is in equilibrium if and only if the total

virtual work of external forces acting on the body is

zero for any virtual displacement of the body.

• If a system of connected rigid bodies remains connected

during the virtual displacement, only the work of the

external forces need be considered.

11

© 2007 The McGraw-Hill Companies, Inc. All rights reserved.

V

ector Mechanics for Engineers: Statics

Eighth

Edition

3 - 21

Sample Problem 10.1

Determine the magnitude of the couple Mrequired to

maintain the equilibrium of the mechanism.

SOLUTION:

• Apply the principle of virtual work

D

PM

xPM

UUU

δδθ

δ

δ

δ

+=

+

=

=

0

0

θδθδ

θ

sin3

cos3

lx

lx

D

D

−=

=

(

)

θ

δ

θ

δθ

sin30 lP

M

−

+

=

θ

sin3Pl

M

=

© 2007 The McGraw-Hill Companies, Inc. All rights reserved.

V

ector Mechanics for Engineers: Statics

Eighth

Edition

3 - 22

Potential Energy and Equilibrium

(not covered in the final)

• When the potential energy of a system is known,

the principle of virtual work becomes

θ

δθ

θ

δδ

d

dV

d

dV

VU

=

−=−==

0

0

• For the structure shown,

( ) ( )

θθ cossin2

2

2

1

2

2

1

lWlk

WykxVVV

CBge

+=

+=+=

• At the position of equilibrium,

( )

Wkll

d

dV

−==

θθ

θ

cos4sin0

indicating two positions of equilibrium.

12

© 2007 The McGraw-Hill Companies, Inc. All rights reserved.

V

ector Mechanics for Engineers: Statics

Eighth

Edition

3 - 23

Stability of Equilibrium

(not covered in the final)

0

2

2

>

θ

d

Vd

2

2

0

d V

dθ

<

0=

θ

d

dV

Must examine higher

order derivatives.

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